Calculating Energy from a Kilogram of Fissionable Fuel

The total energy produced is the number of ${}^{\text{235}}\text{U}$ atoms times the given energy per ${}^{\text{235}}\text{U}$ fission. We should therefore find the number of ${}^{\text{235}}\text{U}$ atoms in 1.00 kg.

The number of ${}^{\text{235}}\text{U}$ atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of ${}^{\text{235}}\text{U}$ has a mass of 235.04 g; thus, there are $(\text{1000 g})/(\text{235.04 g/mol})=\mathrm{4.25\; mol}$. The number of ${}^{\text{235}}\text{U}$ atoms is therefore,

So the total energy released is

This is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only one-fourth the energy produced by the fusion of a kilogram mixture of deuterium and tritium as seen in Example 32.2. Even though each fission reaction yields about ten times the energy of a fusion reaction, the energy per kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much more scarce than fusion fuel, and less than 1% of uranium $\text{(the}{}^{\text{235}}\text{U}\text{)}$ is readily usable.